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Answer $(1,1,1)$, $(1,2,2)$, $(2,3,6)$. Let $p=a+b+c$, then $a+b=p-c$, $b+c=p-a$, $c+a=p-b$ and $$\frac{(p-c{)}^4}{c}+\frac{(p-a{)}^4}{a}+\frac{(p-b{)}^4}{b}$$ is a non-negative integer. By expanding brackets we obtain that the number $p^4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is integer, too. But the numbers $a$, $b$, $c$ are not divisible by $p$, therefore the number $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is (non negative) integer. That is possible for the triples $(1,1,1)$, $(1,2,2)$, $(2,3,6)$ only.